Nice problem. I'll submit my solution after a while.

Easier than last year's #6. Let $P', Q'$ be the reflections of $A$ over $P, Q$ respectively. Let $T = BC \cap PQ.$ First of all, notice that since $AD = BC = AE$, we've that $\triangle ADE$ is isosceles. Hence, $AM$ must pass through $O$, the circumcenter of $\triangle ABC.$ Consequently, we have that $Y = A'$, the point diametrically opposite $A$ on $\Gamma.$ Also, we get $AJ \perp PQ.$ Since $QA = QB$, we have that $QA = QB = QQ' \Rightarrow \angle Q'BA = 90$. Analogously, $\angle P'CA = 90.$ Therefore, $Y = BQ' \cap CP'$, and so by Pappus' Theorem on degenerate hexagon $BQP'CPQ'$ we get that $X, Y, PQ' \cap QP'$ are collinear. As $PQ' \cap QP'$ is the centroid of $\triangle AP'Q'$, and $Y$ is the orthocenter of $\triangle AP'Q'$, it follows that $X, Y, PQ' \cap QP'$ are actually collinear on the Euler line of $\triangle AP'Q'.$ Therefore, it suffices only to show that $Z$ is on the Euler line of $\triangle AP'Q'.$ Let $N$ be the circumcenter of $BQCP$, i.e., the nine-point center of $\triangle AP'Q'.$ By Brokard's Theorem on $BQCP$, it follows that $NX \perp AT.$ Hence, as $N$ is clearly on the Euler line of $\triangle AP'Q'$ (a.k.a. $XY$), we just need to show that $YZ \perp AT.$ Since $AY \perp TZ$, we want to prove that $Y$ is the orthocenter of $\triangle AZT.$ Since $AJ \perp TZ$, we only need to show that $JY \cdot JA = JT \cdot JZ.$ We have by Power of the Point that $$JT \cdot JZ = TJ \cdot TZ - TJ^2 = TB \cdot TC - TJ^2 = TO^2 - R^2 - TJ^2,$$where $R$ is the circumradius of $\triangle ABC.$ Also, we have that $$JY \cdot JA = (JO - OY)(JO+OA) = JO^2 - R^2.$$Hence, we only need to show that: $$TO^2 - R^2 - TJ^2 = JO^2 - R^2 \Leftrightarrow TO^2 = TJ^2 + JO^2,$$but this is obvious by the Pythagorean Theorem applied on $\triangle TJO.$ $\square$

Solution. Since $ABDC$ and $ABCE$ are isosceles trapezoids, we have $AD=BC=AE$, thus $AM$ is the perpendicular bisector of $DE$, so $AM\perp DE$ and $Y$ is the antipode of $A$ on $\Gamma$. Observe that triangles $BPD$ and $BQA$ are directly similar, i.e. $B$ maps $\overline{PD}$ to $\overline{QA}$, thus it also takes $\overline{PQ}$ to $\overline{DA}$, in other words, $\bigtriangleup ABD\sim \bigtriangleup QBP$, therefore $$\angle BQP=\angle BAD=\angle BED=\angle BCP$$implying that $BQCP$ is cyclic and $PQ\parallel DE$, so $AJ\perp PQ$. Because $\angle ACY=\angle ABY=90^\circ$, the latter leads to infer that $QCYJ$ and $PBJY$ are cyclic. Let $T=\overline{BC}\cap \overline{PQ}$ and $R=\overline{AT}\cap \Gamma,\ R\neq A$. We have $\angle YRT=\angle TJY=90^\circ$, so $TJYR$ is cyclic. Then $$AT\cdot AR= AJ\cdot AY= AQ\cdot AC= AB\cdot AP $$which yields that $TQCR$ and $PBTR$ are cyclic quadrilaterals; hence $$\angle TRB=\angle TPB=\angle TCQ=\angle TRQ$$i.e. $RA$ is the internal bisector of $\angle BRQ$. Together with $(X,T;B,Q)\overset{R}{=}-1$, we conclude that $\angle ARX=90^\circ$, therefore $X,R$ and $Y$ are collinear. Finally, let $Z'=\overline{TJ}\cap \overline{YR}$ and $S=\overline{YT}\cap \overline{AZ'}$. Since $T$ is the orthocenter of $\bigtriangleup AYZ'$ we deduce that $\angle ASY=90^\circ$, implying that $S$ lies on $\Gamma$. It is clear that $Z'SJY$ is cyclic, therefore $$Z'T\cdot TJ = ST \cdot TY= BT\cdot TC$$thus $BQCZ'$ is cyclic, whence $Z=Z'$. We are done! $\blacksquare$

[asy][asy] unitsize(1.5inches); pair A=dir(100); pair B=dir(221); pair C=dir(-41); pair Y=-A; pair U=(A+C)/2; pair V=(A+B)/2; pair P=extension(U,0,A,B); pair Q=extension(V,0,A,C); pair J=extension(A,Y,P,Q); pair X=extension(B,Q,C,P); pair T=extension(P,Q,B,C); pair Z=extension(Y,X,P,Q); pair D=2*foot(0,P,C)-C; pair E=2*foot(0,B,Q)-B; pair L=extension(Y,B,P,Q); pair K=extension(Y,C,P,Q); pair LL=(A+L)/2; pair KK=(A+K)/2; draw(circumcircle(A,B,C)); draw(A--B--C--cycle); draw(Y--K); draw(Y--L); draw(T--B); draw(K--T); draw(C--D); draw(B--E); draw(Q--LL); draw(P--KK); draw(A--Y); draw(Y--Z); draw(A--L,dotted); draw(A--K,dotted); dot("$A$",A,dir(A)); dot("$B$",B,dir(250)); dot("$C$",C,dir(270)); dot("$Y$",Y,dir(Y)); dot("$T$",T,dir(180)); dot("$P$",P,dir(110)); dot("$Q$",Q,dir(80)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$U$",U,dir(75)); dot("$V$",V,dir(110)); dot("$L$",L,dir(120)); dot("$K$",K,dir(K)); dot("$J$",J,dir(59)); dot("$X$",X,dir(250)); dot("$O$",0,dir(245)); dot("$L'$",LL,dir(100)); dot("$K'$",KK,dir(70)); dot("$Z$",Z,dir(60)); [/asy][/asy] Redefine $Z$ to be $XY\cap PQ$. We'll show that $BZJC$ is cyclic. Let $U$ be the midpoint of $AC$ and $V$ the midpoint of $AB$. Note that $ADBC$ and $AECB$ are isosceles trapezoids, so $PU$ and $QV$ are the perpendicular bisectors of $AC$ and $AB$ respectively, so in particular they pass through the circumcenter $O$ of $(ABC)$. Furthermore, \[\widehat{AD}=\widehat{BC}=\widehat{AE},\]so the line $AM$ passes through $O$, so in particular $Y$ is the antipode of $A$ (we won't need $M$, $D$, or $E$ anymore). Note that $O$ is the orthocenter of $APQ$ with altitudes $PU$ and $QV$, so in fact $AJ$ is also an altitude. Thus, we see that $(UVPQ)$ is cyclic, and since $UV\parallel BC$, by Reim's theorem, $(PQBC)$ is also cyclic. Letting $T=PQ\cap BC$, we see then that $TB\cdot TC=TP\cdot TQ$, so to show $BCJZ$ cyclic it suffices to show that $TP\cdot TQ=TZ\cdot TJ$. Define $L=YB\cap PQ$ and $K=YC\cap PQ$. We claim that $(BCLK)$ is cyclic. To see this, perform a homothety at $A$ with scale factor $1/2$. This ssends $BYC\mapsto VOU$, so $L$ and $K$ go to the intersections of $OV$ and $OU$ with the perpendicular bisector of $AJ$. Call these images $L'$ and $K'$. Note that $L'K'\parallel PQ$ and $(UVPQ)$ cyclic, so by Reim's theorem, we see that $(L'K'UV)$ is cyclic. Reversing the homothety, this implies that $(LKBC)$ is cyclic. Therefore, we have $TL\cdot TK=TP\cdot TQ$, so $(P,Q)$ and $(L,K)$ are swapped under the involution $\phi_T$ given by inversion at $T$ with power $TL\cdot TK=TP\cdot TQ$. By DDIT on $PQBC$, there is a projective involution on the pencil of lines through $Y$ swapping the pairs $(YP,YQ)$, $(YB,YC)$, and $(YX,YA)$. Projecting onto line $PQ$, we see that there is an involution swapping $(P,Q)$, $(L,K)$, and $(Z,J)$. By the above observation, this involution must be $\phi_T$, so we have \[TZ\cdot TJ=TP\cdot TQ=TB\cdot TC,\]as desired.

Here's a simple solution. First get $\angle BPC = \angle BQC = 2 \angle A$, so $BPQC$ is cyclic. Thus by Reim's, $PQ \parallel DE$. Angle chase to get $AD = AE$, so $AM \perp DE$ and thus $AM \perp PQ$. As $BPQC$ is cyclic it follows that $Y$ is the $A$-antipode. Let $YX$ cut $PQ$ at $Z'$, so it suffices to prove $Z' \in (BCJ)$. Now let $R$ be the center of spiral similarity from $PQ$ to $BC$, which lies on $(ABC)$. Then $R$ is the Miquel Point of cyclic quadrilateral $BPQC$, so by known properties of the Miquel Point of a cyclic quadrilateral we have $AR, PQ, BC$ concurrent at a point $T$, and $\angle XRA = 90^{\circ}$. The latter implies that $R, X, Y$ are collinear, since $\angle YRA = 90^{\circ}$ too. Finally, remark that $ARZ'J$ is cyclic as angles $ARZ'$ and $AJZ'$ are both right. Thus by Power of a Point from $T$ we have $$TB \cdot TC = TR \cdot TA = TZ' \cdot TJ$$ So $BZ'JC$ is cyclic, as desired.

Why $\angle XRA=90^{\circ}$ ? Explain me, please

Surprisingly, this problem has a hidden Euler Line configuration. We will state the configuration first and use to destroy the problem afterwards. Lemma: Let $ABC$ be a triangle with circumcenter $O$, orthocenter $H$, altitudes $AD, BE, CF$. Let $M_a, M_b,M_c, M$ be midpoints of $BC, CA, AB, AD$. Then we have the following (a) $EM_c, FM_b, OH$ are concurrent. (b) If $T = EF\cap M_bM_c$, then $AT\perp OH$. (c) If $P = OH\cap M_bM_c$, then $E,F,P,M$ are concyclic. Note: Part (a) and (b) are not new. I have seen it somewhere else before attempting this problem. Proof: For (a), apply Pappus theorem on $BM_cF$ and $CM_bE$ to get that $G, H, EM_c\cap FM_b$ are colinear so we are done. (b) also follows quickly by applying Brokard's theorem on $EFM_bM_c$. For (c), notice that $T$ is orthocenter of $\triangle AHP$ thus if $K$ is the projection of $A$ onto $OH$, then $A, M_b, M_c, O, K$ are concyclic. Thus $$TP\cdot TM = TA\cdot TK = TM_b\cdot TM_c = TE\cdot TF$$thus $E,F,M,P$ are concyclic as desired. Back to the main problem. Let $O$ be the circumcenter of $\triangle ABC$. Note that $AD=AE=BC$ thus $AO$ is the perpendicular bisector of $DE$. Therefore $A,O,M$ are colinear $\implies$ $Y$ is the antipode of $A$. Let $P', Q'$ be the image of $P,Q$ under homothety $\mathcal{H}(A,2)$. Notice that as $OP$ is perpendicular bisector of $AC$, we get $OP\perp AC$ $\implies$ $P'Y\perp AC$. Thus $B\in P'Y$. Similarly, $C\in Q'Y$ thus $Y$ is orthocenter of $\triangle AP'Q'$. Now, apply the lemma on $\triangle AP'Q'$. We see that (a) implies $X$ lie on Euler Line of $\triangle AP'Q'$. Moreover, (c) implies $Z$ lie on this line too so we are done.

Let $K=PQ\cap BC$, $R$ the circumcenter of $\triangle BOC$, $T=XK\cap AR$ and $L=AK\cap (ABC)$. 1- notice that $AE=AD$, then $O\in AY$ and $AO\perp DE$. 2- spiral similarity centered at $B$ sending $(P,D)$ to $(Q,A)$ shows that $\angle BQD=\angle BAD=\angle BED$, then $PQ\parallel DE$ and $PQ\perp AO$ at $J$. 3- Easy to show that $BPCQO$ is cyclic and by Brokard $\angle XTA=90$. 4-inversion $\phi$ centered at $A$ which fix $(BPCQO)$ sends the line $PQ$ to $(ABC)$, then $\phi$ sends $(K,T,J)$ to $(L,R,Y)$. Using that $AKTJ$ is cyclic we have thet $L,R,Y$ belongs to the same line $\ell$, actually by Brocard $X\in \ell$. 5-by power of point $ZK\cdot KJ=BK\cdot KC=AK\cdot KL$ then $AJLZ$ is cyclic, then $LZ\perp LA$ and this means $Z\in \ell$.

Proof: Firstly notice that $ADBC , AECB$ are cyclic trapezoids $\implies AD=AE \implies AB$ is angle bisector of $\angle DBE$. Now notice that since $\angle BDP=\angle BAQ \implies \Delta BDP \stackrel{+}{\sim} \Delta BAQ \implies PQBC$ is cyclic by spiral similiarity.Now notice that $\angle QPC=\angle QBC=\angle EDC \implies DE \parallel PQ \implies AJ \perp PQ \implies Y$ is the $A$-antipode w.r.t $\odot(ABC)$. Now let $PQ \cap BC=T$ and let $K$ be the miquel point of $PQBC$. Its well known that $\{X,K\}$ are inverses of each other w.r.t $\odot(PQBC) \implies \{A,K,T\}$ is a collinear triple by brocards theorem on $PQBC$. Also we have $\angle XKA=90^\circ$. But by definition we have $K \in \odot(ABC) \implies \angle YKA=90^\circ \implies \{Y,X,K\}$ is a collinear triple. Now notice that by radical axis theorem on$ \odot(BZJC),\odot(AKZ),\odot(ABC)$ we have that $J\in \odot(AKZ) \implies \angle AKZ=90^\circ$. But since $\angle AKY=90^\circ \implies \{Y,K,Z\}$ is a collinear triplet $\implies \{Y,X,Z,K\}$ is a collinear tuple $\implies \{Y,X,Z\}$ collinear as desired $\blacksquare$. 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Lol this is what happens when you don't solve easy problems for a long time. Jafet98 wrote: Let $\Gamma$ be the circumcircle of triangle $ABC$. The line parallel to $AC$ passing through $B$ meets $\Gamma$ at $D$ ($D\neq B$), and the line parallel to $AB$ passing through $C$ intersects $\Gamma$ to $E$ ($E\neq C$). Lines $AB$ and $CD$ meet at $P$, and lines $AC$ and $BE$ meet at $Q$. Let $M$ be the midpoint of $DE$. Line $AM$ meets $\Gamma$ at $Y$ ($Y\neq A$) and line $PQ$ at $J$. Line $PQ$ intersects the circumcircle of triangle $BCJ$ at $Z$ ($Z\neq J$). If lines $BQ$ and $CP$ meet each other at $X$, show that $X$ lies on the line $YZ$. $\textbf{LEMMA:-}$ $\ell_1$ and $\ell_2$ be two lines intersecting at $X$. Let $\{P,Q\}\in\ell_1$ and $\{R,S\}\in\ell_2$ such that $\overline{PR},\overline{QS}$ are antiparallel sides. Let $\overline{PR}\cap\overline{QS}=V$. Let $A,B$ be the $X-$ Antipodes WRT $\Delta XQR$ and $\Delta XPS$. Then $\overline{V-A-B}$ are collinear. Animate $R$ on $\ell_2$. So, $R\mapsto Q$ is a Homography between $\ell_1$ and $\ell_2$. Also $R\mapsto Q\mapsto \overline{R\infty_{XS}}\mapsto\overline{Q\infty_{XP}}$ are Homographies. So, $\deg(Q\infty_{XP})=\deg(R\infty_{XS})=1\implies\deg(A)\leq 1+1=2$. $\deg(V)\le\deg(PR)+\deg(QS)=1+1=2$. So in order to prove $V,A,B$ collinear it suffices to check $\deg(A)+\deg(V)+\deg(B)+1=2+2+0+1=5$ cases for the Position of $R$ on $\ell_2$. When $R=S,X$ and $\overline{PR}\perp\overline{XS}$ and when $Q=P$ the result is trivial. When $\overline{PR}\|\overline{QS}$ then $V=\infty_{QS}$ and $\overline{AB}\|\overline{QS}$ which implies $V,A,B$ are collinear. So with this we prove our Lemma. Let $\overline{XY}\cap\odot(ABC)=F$. Notice that $AD=BC=AE\implies \overline{AY}\perp\overline{DE}$ and $\measuredangle AYB=\measuredangle ACB=\measuredangle QPA\implies Y$ is the $A-$ antipode in $\odot(ABC)$. Let $U$ be the $A-$ antipode in $\Delta APQ$. So by our $\textbf{LEMMA}$ we get $Y,X,U,F$ are collinear $\implies F\in\odot(APQ)$ so $F$ is the Miquel Point of the Complete Quadrilateral formed by the lines $\{\overline{BP},\overline{PQ},\overline{QC},\overline{CB}\}$. Now by applying Radical Axis Theorem on $\odot(AFPQ),\odot(BCPQ),\odot(ABC)$ we get $\overline{AF},\overline{PQ},\overline{BC}$ are concurrent at $K$. Let $\overline{YF}\cap\overline{PQ}=Z^*$. Then $AJZ^*F$ is a cyclic quadrilateral. So, $KZ^*\cdot KJ=KB\cdot KC\implies Z^*\equiv Z$. $\blacksquare$

Nice problem. Clearly, points $P,Q$ can be defined as the points on $AB$ and $AC$, respectively, such that $PA=PC$ and $QA=QB$. From now on, we completely ignore points $D,E$. Also, we have $\angle ADE=\angle AED=\angle A$, hence $AD=AE$, so $AM \perp DE$. In addition, $BPQC$ is cyclic, since $\angle ABQ=\angle ACP=\angle A$. Hence, $\angle APQ=\angle C \Rightarrow \angle BAY=90^\circ-\angle C$, implying that $Y$ is the antidiametrical point of $A$. We restate now the problem (changing the labelling as well): Restated problem wrote: Let $\Gamma$ be the circumcircle of triangle $ABC$, and $A'$ be the antidiametrical point of $A$ on this circle. Let $P,Q$ be on $AB,AC$ respectively, such that $PA=PC$ and $QA=QB$. Let $BQ$ and $CP$ meet at point $X$. Line $AA'$ intersects $PQ$ at $T$, and line $A'X$ intersects $PQ$ at $S$. Then $BSTC$ is cyclic. Let $O$ be the circumcenter of the quadrilateral $BPQC$. We prove the following Claim: Claim: Points $X,O,A'$ are collinear. Proof: Let the projections of $X,O$ on $AB$ be $K,L$, respectively, and on $AC$ be $M,N$. It suffices to show that $\dfrac{BK}{BL}=\dfrac{MC}{CN}$. But, this is trivial, since triangles $\vartriangle BPX, \vartriangle XQC$ are similar, so $\dfrac{BK}{BP}=\dfrac{MC}{CQ}$, which gives the desired relation, using that $BK=2BL, CQ=2CN$ $\blacksquare$. Back to the problem. Let $R \equiv PQ \cap BC$, and let $AR$ intersect $(\Gamma)$ at $U$. Then, by Brocard's, $OX \perp AR$. In addition, $A'U \perp AR$. So, using the Claim, we obtain that $U,X,O,A'$ are collinear. To end, we have that $\angle SUA=\angle STA=90^\circ$, so $SUAT$ is cyclic. Hence, $RS \cdot RT=RU \cdot RA=RB \cdot RC$, which implies that $BSTC$ is cyclic, hence we are done.

This is my solution using Pascal theorem, radical axis theorem and inversion at $A$

Let $W=BC\cap PQ$, $S=AW\cap (ABC), Z'=AZ\cap (ABC)$. Trivially $AD=BC=AE$, so $AM\perp DE$, and so $Y$ is the antipode of $A$ in $(ABC)$. Also, $$ \measuredangle PCQ=\measuredangle DBA=\measuredangle ACE=\measuredangle PBQ \Longrightarrow PBCQ \thickspace \text{cyclic.} $$Hence $$ \frac{XP}{XQ}=\frac{XB}{XC}=\frac{XD}{XE} \Longrightarrow PQ\parallel DE. $$Claim 1. $S-Y-Z$ collinear. Proof. Let $\psi$ be the inversion around $A$ of radius $\sqrt{AB\cdot AP}$. Clearly, $\psi$ maps $B\rightarrow P ; C\rightarrow Q; S\rightarrow W; Y\rightarrow J, Z\rightarrow Z'$. Now, $$ WJ\cdot WZ=WC\cdot WB=WS\cdot WA \Longrightarrow ASJZ \thickspace \text{cyclic.} $$Therefore, $\psi(S), \psi(J), \psi(Z)$ are collinear, which means $W-Y-Z'$ collinear. In triangle $\triangle AWZ$, $AJ$ and $WZ'$ are altitudes, so $Y$ is its orthocenter, and so $ZY\perp AW$, which means $Z-Y-S$ collinear. Claim 2. $S-X-Y$ collinear. Proof. Since $BQ\cap CP=X$, then $$ (E,D;A,Y)=-1=(P,Q;AX\cap PQ, W)\stackrel{(A)}=(B,C,AX\cap (ABC), S)\stackrel{(X)}=(E,D;A,SX\cap (ABC)), $$which means $S-X-Y$ are collinear. Both claims finish the problem.

$\textbf{Claim 1.}$ $Y$ is the antipode of $A.$ $\textit{Proof.}$ Computing angle $\measuredangle (DE,BC)$ one easely gets that $DE$ is antiparallel to $BC$ w.r.t. $AB$ and $AC.$ Thus $AM\perp DE $ shows the claim. $\hfill\blacksquare$ $\textbf{Claim 2.}$ $BQCP$ ciclic. $\textit{Proof.}$ We apply Pascal at $AABEDC,$ and obtain that $AA\cap DE - P - Q,$ but we showed that $DE$ antiparallel, which implies that $AA\cap BC=\infty_{DE},$ implying that $PQ$ is also antiparallel, showing the ciclic. $\hfill\blacksquare$ Now let $K=PQ\cap BC$ and $T=AK\cap (ABC).$ Now, $KA\cdot KT=KB\cdot KC=KA\cdot KT \implies AJTZ$ ciclic. But as $AJ\perp PQ,$ then $AT\perp ZT \implies Z-T-Y.$ But now Brokard in $BQCP$ tells us that the problem is equivalent to $Z-T-Y$ going through the circumcenter of $BQCP.$ But this is well-known: $\textbf{Well-known}$ Let $ABCD$ be ciclic with center $O,$ and $K=AD\cap BC,$ $P=AC\cap BD,$ $Q=AB\cap CD.$ We now by Brokard that $K$ is the orthocenter of $OPQ.$ Now, the inversion at $K$ with negative power which swaps $O,P,Q$ with their feet in $\triangle OPQ,$ lets $ABCD$ invariable.

To better follow along, assume $AB < BC < AC$, thus $D$ and $A$ lie on opposite sides wrt $BC$, while $E$ and $A$ lie on the same side of $BC$. Similar to other solutions we can note that $BPCQ$ is cyclic and $AY$ is the perpendicular bisector of $DE$. For the former note that \[ \angle QBA = \angle BAC = \angle ACP. \]For the latter note that $AE = BC = AD$, and since $M$ is the midpoint of $DE$ we conclude $AY$ is the perpendicular bisector of $DE$. Now note that $PQ \parallel DE$, for \[ \angle CDE = \angle CBE = \angle CBQ = \angle CPQ. \]In particular, this implies $AY \perp PQ$. Note now that $JQCY$ is cyclic, and furthermore \[ AJ \cdot AY = AQ \cdot AC = AB \cdot AP \]thus $BPYJ$ is cyclic as well. Note now that \begin{align*} \angle DCZ &= \angle BCZ - \angle BCD \\ &= \angle BJZ - \angle BCD \\ &= \angle BJP - \angle BCD \\ &= \angle BYP - \angle BCD \\ &= \angle BYD + \angle DYP - \angle BCD \\ &= \angle DYP. \end{align*}This implies that $CZ$ and $PY$ meet at a point $R$ in $\Gamma$. Similarly, $BZ$ and $QY$ meet at a point $S$ in $\Gamma$. From here we can finish with Menelaus and a few similarities. For, let $BQ$ meet $ZY$ at $X_1$ and let $CP$ meet $ZY$ at $X_2$. By Menelaus on $SZY$ and $RZY$ wrt lines $Q-B-X_1$ and $C-P-X_2$ we get \[ \frac{SQ}{QY} \cdot \frac{YX_1}{X_2Z} \cdot \frac{ZB}{BS} = 1 = \frac{RP}{RY} \cdot \frac{YX_2}{X_2Z} \cdot \frac{ZC}{CR}. \]Thus we have to prove that \[ \frac{SQ}{QY} \cdot \frac{ZB}{BS} = \frac{RP}{RY} \cdot \frac{ZC}{CR}. \]To achieve this we might consider the following similar triangles and be done.